Answer to Question #139003 in Trigonometry for riz

Question #139003

Find the remaining parts of a quadrantal triangle (c = 90°).

1.) a = 70°10’ b = 52°40’

2.) a = 116°53’ A = 122°39’

3.) b = 69°29.7’ B = 63°4.6’

4.) a = 106°38’ b = 36°49’

5.) A = 52°55’ b = 73°11’


1
Expert's answer
2020-10-22T18:02:14-0400

In case giving a spherical triangle in which one side is 90° , we use the following fundamental formuleas




"\\sin middle=\\tan(adj.)\\times\\tan(adj.)""\\sin middle=\\cos(opp.)\\times\\cos(opp.)"

1.) a = 70°10’ b = 52°40’ c = 90°


"\\sin (-(90\\degree-C))=\\tan(90\\degree- b)\\times \\tan(90\\degree-a)""\\cos C=-\\cot a\\times\\cot b""C=\\cos^{-1}(-\\cot(70\\degree10')\\cot(52\\degree40')\\approx105\\degree58'4''""\\sin (90\\degree-b)=\\cos B\\times \\cos(90\\degree-a)""\\cos B=\\dfrac{\\cos b}{\\sin a}""B=\\cos^{-1}(\\dfrac{\\cos 52\\degree40'}{\\sin 70\\degree10'})\\approx49\\degree51'27''""\\sin (90\\degree-a)=\\cos A\\times \\cos(90\\degree-b)""\\cos A=\\dfrac{\\cos a}{\\sin b}""A=\\cos^{-1}(\\dfrac{\\cos 70\\degree10'}{\\sin 52\\degree40'})\\approx64\\degree44'28''"

2.) a = 116°53’ A = 122°39’ c = 90°



"\\sin (90\\degree-a)=\\cos A\\times \\cos(90\\degree-b)""\\sin b=\\dfrac{\\cos a}{\\cos A}""b=\\sin^{-1}(\\dfrac{\\cos 116\\degree53'}{\\cos 122\\degree39'})\\approx56\\degree56'35''"




"\\sin B=\\tan A\\times \\tan (90\\degree-a)""B=\\sin^{-1}(\\dfrac{\\tan 122\\degree39'}{\\tan116\\degree53'}\\ )\\approx52\\degree17'50''""\\sin (-(90\\degree-C))=\\tan (90\\degree-b)\\times \\tan (90\\degree-a)""\\cos C=-\\dfrac{1}{\\tan a\\times\\tan b}""C=\\cos^{-1}(-\\dfrac{1}{\\tan 116\\degree53'\\times\\tan 122\\degree39'})\\approx108\\degree57'21''"


3.) b = 69°29.7’ B = 63°4.6’ c = 90°



"\\sin A=\\tan B\\times \\tan (90\\degree-b)""A=\\sin^{-1}(\\dfrac{\\tan63\\degree4.6'}{\\tan69\\degree29.7'})\\approx47\\degree25.6'""\\sin (90\\degree-b)=\\cos B\\times \\cos(90\\degree-a)""\\sin a=\\dfrac{\\cos b}{\\cos B}""a=\\sin^{-1}(\\dfrac{\\cos69\\degree29.7'}{\\cos63\\degree4.6'})\\approx50\\degree40.8'""\\sin B=\\cos (90\\degree-b)\\times \\cos(-(90\\degree-C))""\\sin C=\\dfrac{\\sin B}{\\sin b}""C=\\sin^{-1}(\\dfrac{\\sin63\\degree4.6'}{\\sin69\\degree29.7'})\\approx72\\degree9.7'"

4.) a = 106°38’ b = 36°49’ c = 90°



"\\sin (-(90\\degree-C))=\\tan(90\\degree- b)\\times \\tan(90\\degree-a)""\\cos C=-\\cot a\\times\\cot b""C=\\cos^{-1}(-\\cot(106\\degree38')\\cot(36\\degree49')\\approx66\\degree28'41''""\\sin (90\\degree-b)=\\cos B\\times \\cos(90\\degree-a)""\\cos B=\\dfrac{\\cos b}{\\sin a}""B=\\cos^{-1}(\\dfrac{\\cos 36\\degree49'}{\\sin 106\\degree38'})\\approx49\\degree51'27''""\\sin (90\\degree-a)=\\cos A\\times \\cos(90\\degree-b)""\\cos A=\\dfrac{\\cos a}{\\sin b}""A=\\cos^{-1}(\\dfrac{\\cos 106\\degree38'}{\\sin 36\\degree49'})\\approx118\\degree32'"



5.) A = 52°55’ b = 73°11’



"\\sin A=\\tan B\\times \\tan (90\\degree-b)""\\tan B=\\sin A\\times\\tan b""B=\\tan^{-1}(\\sin52\\degree55'\\times\\tan73\\degree11')\\approx69\\degree15'2''""\\sin (90\\degree-a)=\\cos A\\times \\cos(90\\degree-b)""a=\\cos^{-1}(\\cos52\\degree55'\\times\\sin73\\degree11')\\approx54\\degree44'49'""\\sin (90\\degree-b)=\\tan (-(90\\degree-C))\\times \\tan A""\\tan C=-\\dfrac{\\tan A}{\\cos b}""C=180\\degree-\\tan^{-1}(\\dfrac{\\tan52\\degree55'}{\\cos73\\degree11'})\\approx102\\degree20'5''"

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